Q52 / 120 · VITEEE 2017 · Mathematics

The value of $x$ obtained from the equation $\left|\begin{array}{ccc}x+\alpha & \beta & \gamma \\ \gamma & x+\beta & \alpha \\ \alpha & \beta & x+\gamma\end{array}\right|=0$ will be

  • Correct answer
    A
    0 and $-(\alpha+\beta+\gamma)$
  • B
    0 and $\alpha+\beta+\gamma$
  • C
    1 and $(\alpha-\beta-\gamma)$
  • D
    0 and $\alpha^{2}+\beta^{2}+\gamma^{2}$

Solution

Given $\left|\begin{array}{ccc}x+\alpha & \beta & \gamma \\ \gamma & x+\beta & \alpha \\ \alpha & \beta & x+\gamma\end{array}\right|=0$ Operate $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$ $\begin{array}{l} \left|\begin{array}{ccc} x+\alpha+\beta+\gamma & \beta & \gamma \\ x+\alpha+\beta+\gamma & x+\beta & \alpha \\ x+\alpha+\beta+\gamma & \beta & x+\gamma \end{array}\right|=0 \\ =(x+\alpha+\beta+\gamma)\left|\begin{array}{ccc} 1 & \beta & \gamma \\ 1 & x+\beta & \alpha \\ 1 & \beta & x+\gamma \end{array}\right|=0 \\ \Rightarrow x+\alpha+\beta+\gamma=0 \Rightarrow x=-(\alpha+\beta+\gamma) \end{array}$ Again if $$ \begin{array}{l} \left|\begin{array}{ccc} 1 & \beta & \gamma \\ 1 & x+\beta & \alpha \\ 1 & \beta & \gamma \end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc} 1 & \beta & \gamma \\ 0 & x & \alpha-\gamma \\ 0 & 0 & x \end{array}\right|=0 \\ \Rightarrow x^{2}=0 \Rightarrow x=0 \end{array} $$ $\therefore$ Solutions of the equation are $x=0$, $-(\alpha+\beta+\gamma)$

Hence, the correct answer is option: 'A'.

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